Can someone help me with this proof for the sum of the first n cubes?

Question

I am trying to prove that $\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4} = \frac{n^4+2n^3+n^2}{4}$.

I experimented by setting n = 5 and writing $1 + 8 + 27 + 64 + 125 = 1 + (1 + 7) + (1 + 7 + 19) + (1 + 7 + 19 + 37) + (1 + 7 + 19 + 37 + 61)$

We can rewrite this as $1 + (1 + (1 + 6)) + (1 + (1 + 6) + (1 + 6 + 12)) + (1 + (1 + 6) + (1 + 6 + 12) + (1 + 6 + 12 + 18)) + (1 + (1 + 6) + (1 + 6 + 12) + (1 + 6 + 12 + 18) + (1 + 6 + 12 + 18 + 24))$.

This is $\sum_{i=1}^5 i^3 = \sum_{i=1}^5 i + 6\sum_{i=1}^4 i + 12\sum_{i=1}^3 i + + 18\sum_{i=1}^2 i + 24$.

For any n, we write $\sum_{i=1}^n i^3 = \sum_{i=1}^n i + \sum_{i=1}^{n - 1}[6i\sum_{j=1}^{n - i}j]$.

Carrying out some summations, we get $\sum_{i=1}^n i^3 = \frac{n(n+1)}{2} + \sum_{i=1}^{n - 1}\frac{6i(n-i)(n-i+1)}{2}$.

Knowing that $i(n-i)(n-i+1)$ expands to $n^2i - ni^2 + ni - ni^2 + i^3 - i^2$, we write:

$\sum_{i=1}^n i^3 = \frac{n(n+1)}{2} + 3[(n^2 + n) \sum_{i=1}^{n - 1}i + (-2n -1)\sum_{i=1}^{n - 1}i^2 + \sum_{i=1}^{n - 1}i^3]$

We rewrite $\sum_{i=1}^{n - 1}i^3$ as $\sum_{i=1}^{n}i^3 - (n-1)^3$

Combining the $i^3$ summations and expanding, we get $-2\sum_{i=1}^{n}i^3 = \frac{n(n+1)}{2} + 3(n^2 + n)\frac{(n-1)n}{2} + 3(-2n-1)\frac{(n-1)(n)(2n-1)}{6} - 3(n-1)^3$

This simplifies to $\sum_{i=1}^{n}i^3 = \frac{n^4+2n^3-17n^2+18n-6}{4}$, which is off from our desired result of $\frac{n^4+2n^3+n^2}{4}$. Any thoughts would be appreciated. Thank you.

Answer 1

Instead of rewriting $\sum\limits_{i=1}^{n - 1}i^3$ as $\left(\sum\limits_{i=1}^{n}i^3\right) - (n-1)^3,$

you should write $\sum\limits_{i=1}^{n - 1}i^3$ as $\left(\sum \limits_{i=1}^{n}i^3\right) - \color{red}n^3$,

and then you'll get the correct answer.

Answer 2

Here is a generalization of this.

If $f_k(n) = \sum_{i=1}^n i^k$ ($f_0(n) = n$) then

$\begin{array}\\ f_k(n+1) &= \sum_{i=1}^{n+1} i^k\\ &= \sum_{i=0}^{n} (i+1)^k\\ &=1+ \sum_{i=1}^{n} (i+1)^k\\ &=1+ \sum_{i=1}^{n} \sum_{j=0}^k \binom{k}{j}i^j\\ &=1+ \sum_{j=0}^k \binom{k}{j}\sum_{i=1}^{n}i^j\\ &=1+ \sum_{j=0}^k \binom{k}{j}f_j(n)\\ \text{so}\\ (n+1)^k &=f_{k}(n+1)-f_k(n)\\ &=1+ \sum_{j=0}^k \binom{k}{j}f_j(n)-f_k(n)\\ &=1+ \sum_{j=0}^{k-1} \binom{k}{j}f_j(n)\\ \end{array} $

or $kf_{k-1}(n) =(n+1)^k-1-\sum_{j=0}^{k-2} \binom{k}{j}f_j(n) $

or $(k+1)f_{k}(n) =(n+1)^{k+1}-1-\sum_{j=0}^{k-1} \binom{k+1}{j}f_j(n) $

Examples:

If $k=1$ then

$\begin{array}\\ 2f_{1}(n) &=(n+1)^{2}-1-f_0(n)\\ &=n^2+2n+1-1-n\\ &=n^2+n\\ \end{array} $

If $k=2$ then

$\begin{array}\\ 3f_{2}(n) &=(n+1)^{3}-1-f_0(n)-3f_1(n)\\ &=n^3+3n^2+3n+1-1-n-3(n^2+n)/2\\ &=\dfrac{2n^3+6n^2+6n+2-2-2n-3(n^2+n)}{2}\\ &=\dfrac{2n^3+3n^2+n}{2}\\ &=\dfrac{n(2n+1)(n+1)}{2}\\ \end{array} $

If $k=3$ then

$\begin{array}\\ 4f_{3}(n) &=(n+1)^{4}-1-f_0(n)-4f_1(n)-6f_2(n)\\ &=n^4+4n^3+6n^2+4n+1-1-n-4\dfrac{n^2+n}{2}-6\dfrac{n(2n+1)(n+1)}{6}\\ &=n^4+4n^3+6n^2+3n-2(n^2+n)-(2n^3+3n^2+n)\\ &=n^4+2n^3+n^2\\ &=n^2(n^2+2n+1)\\ &=n^2(n+1)^2\\ \end{array} $

Answer 3

There is a general method for summing the powers of $i$, which results from the properties of the binomial coefficients.

Indeed we have the formula $\displaystyle\sum\limits_{i=1}^n\binom{i}{k}=\binom{n+1}{k+1}\ $ see for instance this post.

Therefore instead of working in the power basis $1,n,n^2,n^3,\cdots$ you can advantageously work in the binomial basis $1,\binom{n}{1},\binom{n}{2},\binom{n}{3},\cdots$

So let's decompose $n^3$ in this basis !

$n^3-n(n-1)(n-2)=3n^2-2n\ $ and then $\,\,3n^2-2n=3n(n-1)+n$

Therefore $n^3=3!\binom{n}{3}+3\cdot 2!\binom{n}{2}+1!\binom{n}{1}=6\binom{n}{3}+6\binom{n}{2}+\binom{n}{1}$

It is then immediate that :

$\begin{align}\textstyle\sum i^3 &=\textstyle6\binom{n+1}{4}+6\binom{n+1}{3}+\binom{n+1}{2}\\ &=\textstyle\frac 6{24}(n+1)n(n-1)(n-2)+\frac 66(n+1)n(n-1)+\frac 12(n+1)n\\ &=\textstyle\frac 14(n+1)n\Big((n-1)(n-2)+4(n-1)+2\Big)\\ &=\textstyle\frac 14(n+1)n\Big(n^2+n\Big)\\ &=\textstyle\frac 14n^2(n+1)^2 \end{align}$